If the binding energy of `2^(nd)` excited state of a hydrogen like sample os `24 eV` approximately, then the ionisation energy of the sample is approximately

Options

A. `54.4 eV`
B. `24 eV`
C. `122.4 eV`
D. `216 eV`

Explanation

To find the ionization energy of a hydrogen-like atom given the binding energy of its 2nd excited state, we can follow these steps:

### Step 1: Understand the Binding Energy
The binding energy of an electron in a hydrogen-like atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
where:
- \( E_n \) is the binding energy at the nth energy level,
- \( Z \) is the atomic number,
- \( n \) is the principal quantum number (energy level).

### Step 2: Identify the Given Values
From the problem, we know:
- The binding energy of the 2nd excited state (which corresponds to \( n = 3 \)) is \( 24 \, \text{eV} \).
- For hydrogen, \( Z = 1 \).

### Step 3: Set Up the Equation for the 2nd Excited State
Using the binding energy formula for \( n = 3 \):
\[ E_3 = -\frac{13.6 \, \text{eV} \cdot Z^2}{3^2} = -\frac{13.6 \, \text{eV} \cdot 1^2}{9} \]
This simplifies to:
\[ E_3 = -\frac{13.6 \, \text{eV}}{9} \approx -1.51 \, \text{eV} \]
However, since the binding energy is given as \( 24 \, \text{eV} \), we can set up the equation:
\[ -\frac{13.6 \, \text{eV}}{9} = -24 \, \text{eV} \]

### Step 4: Calculate the Ionization Energy
The ionization energy corresponds to the energy required to remove the electron from the ground state (n=1). Using the formula for \( n = 1 \):
\[ E_1 = -\frac{13.6 \, \text{eV} \cdot Z^2}{1^2} = -13.6 \, \text{eV} \]
To find the ionization energy, we need to consider the total energy from the ground state to the point where the electron is completely removed (ionized):
\[ \text{Ionization Energy} = |E_1| = 13.6 \, \text{eV} \]

### Step 5: Relate Ionization Energy to Binding Energy
Since the binding energy of the 2nd excited state is \( 24 \, \text{eV} \), we can use the relationship:
\[ \text{Ionization Energy} = 9 \times \text{Binding Energy of the 2nd excited state} \]
Thus:
\[ \text{Ionization Energy} = 9 \times 24 \, \text{eV} = 216 \, \text{eV} \]

### Final Answer
The ionization energy of the sample is approximately **216 eV**.

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